If the probability of rain on any given day is 50%, what is the probab

Question

If the probability of rain on any given day is 50%, what is the probability that it will rain on at least three days in a row during a five-day period?

A. 3/32
B. 1/4
C. 9/32
D. 5/16
E. 1/2

GMATinsight · Accepted Answer

Interesting question

Here the question says rain on atleast three days  CONSECUTIVELY  
i.e. Exactly three days in row  Days123 or days 234 or days 345 or days 1235 or days 1345 
i.e. Exactly Four days  Days1234 or days 2345 
or exactly 5 days days  12345 
Probability in each case = (1/2)^5

Required Probability = 8*(1/32) = 1/4
Answer: Option B

A VARIATION AND A NEW QUESTION  
   Had the question been Probability of Rain on Atleast three days out of 5

If it rains on exactly 3 days out of 5 then those 3 days can be selected in 5C3 ways and therefore Probability =  5C3*(0.5)^3*(0.5)^2 = 5C3*(1/32)

If it rains on exactly 4 days out of 5 then those 3 days can be selected in 5C4 ways and therefore Probability =  5C4*(0.5)^4*(0.5)^1 = 5C4*(1/32)

If it rains on exactly 5 days out of 5 then those 3 days can be selected in 5C5 ways and therefore Probability =  5C5*(0.5)^5 = 5C3*(1/32)

Total Probability of raining atleast 3 out of 5 days = (1/32)*(5C3+5C4+5C5) = 16/32 = 1/2

PKN · Answer

Say , symbols R=Rain , N=No rain
Given  P(R)=P(N)=\frac{1}{2}

P(at least 3 Rainy days in a   row  )=P(3 rainy days in a row)+P(4 rainy days in a row)+P(5 rainy days in a row)

1) P(3 rainy days in a row)= 5*(\frac{1}{2})^5 
RRRNR
RRRNN
NNRRR
RNRRR
NRRRN

2) P( 4 rainy days in a row)= 2*(\frac{1}{2})^5 
RRRRN
NRRRR

3) P( 5 rainy days in a row)= (\frac{(1}{2)})^5 
RRRRR

So, required probability= 5*(\frac{1}{2})^5 + 2*(\frac{1}{2})^5 + (\frac{1}{2})^5 = 8*(\frac{1}{2})^5 = 8*\frac{1}{32} = \frac{1}{4}

Ans. (B)

chetan2u · Answer

Important not to miss ' in a row'..
At least three days means three cases...
1) 3 days in a row..
So take these three days as ONE R and other two as N
So NNR..
You can place R in 3C1=3 ways

2) 4 days.....4 days in a row OR 3 days in a row and 4th day of rain separated by a non rainy day
So take these three days as ONE R and non-rainy day as N and 4th day as r
So rNR........ rNR, rRN, RNr,rRN
4 ways

3) 5 days in a row
All 5 are one so one way

Total- 3+4+1=8
All possibilities=2*2*2*2*2=32

Probability=8/32=1/4

gmatnovice1122 , what is wrong with your way..
Firstly you are complicating a simple way..
 We use complement when we want to ease our calculations as normal scenario will have more cases 
If talking of complement..
At least three days in a row means MAX two days in a row..
1) two days..
When there are 4 days of rains....RRNRR
When there are three days....RNNRR, NRNRR, ....many ways
When there are 2 days....RRNNN, NRRNN..and so on
2) one day..
4days is not possible
3 days..RNRNR
2 days ...NNRNR, and so on
3) none of the day
NNNNN... One way

So as you see it is more complicated, stick to normal

EgmatQuantExpert · Answer

Solution

Given:  
 • Probability of rain on any given day is 50%

To find:  
 • The probability that it will rain on at least three days in a row during a five-day period

Approach and Working:   
Assume that R represents rain and N represents no rain
As probability of rain is 50%, each of R and N have individual probability ½.

Now, at least 3 rainy days in a row can happen when
 • Exact 3 rain and 2 no rain: RRRNN or NRRRN or NNRRR or RRRNR or RNRRR
• Exact 4 rain and 1 no rain: RRRRN or NRRRR
• Exact 5 rain: RRRRR

Hence, the probability = (5 + 2 + 1) x  (\frac{1}{2})^5  =  \frac{8}{32}  =  \frac{1}{4}

Hence, the correct answer is option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-AT-e1533292152363.png

ScottTargetTestPrep · Answer

Let R = rain and N = no rain. To have at least three days in a row during a five-day period, we could have:

RRRNN, NRRRN, NNRRR, RRRRN, NRRRR, RNRRR, RRRNR and RRRRR.

Notice that the probability of R is ½ and the probability of N is also ½. Thus, the probability of any of the scenarios above is ½ x ½ x ½ x ½ x ½ = (½)^5 = 1/32. Since we have a total of 6 scenarios, the overall probability is 8 x 1/32 = 1/4.

Answer: B

gmatnovice1122 · Answer

chetan2u 
 GMATinsight 
Thanks for your respond. I appreciate the feedback. 
I was confused because when reviewing probabilities, you are thought that when you have the words "AT LEAST" to automatically use the complement. As a result, that is the method I followed. 
Also, wouldn't the complement of  P(Rain on AT LEAST three days in a row) be 1 - P(Rain on AT LEAST two or fewer days in a row.   and NOT as you stated above..so it would be RRNNN,NRRNN, etc? Is my understanding of complements wrong?

In the past I solved a problem related to selecting cards out of a 52 card deck, that stated P (at least the third draw). The complement of this was used 1 - p(two or fewer draws).

Thanks again for helping clarify this.

chetan2u · Answer

Hi..

At least 3 days means on 3 days, 4 days and 5 days..
So complement will be 1-(no days, 1 days and 2 days)
The moment you do at least 2 days, it itself will be contain at least 3 days or at least 4 days..

So if you do 1-(atleast 2 days or few days) means 1-(probability of rain on 2 days, 3 days , 4 days and 5 days)

EMPOWERgmatRichC · Answer

Hi All,

We're told that the probability of rain on any given day is 50%. We're asked for the probability that it will rain on AT LEAST three days in a ROW during a five-day period. This question can be approached in a number of different ways; you might find it easiest to do a little math and 'map out' the various possible ways to get at least 3 days of rain in a row.

Since there are two events (rain or no rain) that have an equal chance of occurring each day, with 5 days in a row, there are (2)(2)(2)(2)(2) = 32 possible outcomes for the 5 days. The number of outcomes that include AT LEAST 3 days of rain IN A ROW are....

Rain on ONLY 3 days in a row: 
RRRNN 
NRRRN 
NNRRR

Rain on 4 of the days (with 3 or 4 days in a row): 
RRRRN 
RRRNR 
RNRRR 
NRRRR

Rain on all 5 of the days: 
RRRRR

8 options out of 32 total = 8/32 = 1/4

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

Abhi077 · Answer

P of raining atleast 3 days out of 5. R is probability it will rain and N is it won't rain 
RRRNN or RRRRN or RRRRR
So you just have to add these probablities.
Probability that it will rain is 1/2. No rain is 1/2
(1/2)^5 + (1/2)^5 +(1/2)^5 =3/32

Posted from my mobile device

kornn · Answer

Dear VeritasKarishma IanStewart EMPOWERgmatRichC chetan2u,

I have tried the complimentary case, i.e. LESS THAN 3 days in a row, and got 24 favorable outcomes as detailed below.

1. 2 days in a row
a) 4 days of rain: RRNRR -> 1 case
b) 3 days of rain: RRNRN RRNNR NRRNR RNRRN RNNRR NRNRR -> 6 cases
c) 2 days of rain: RRNNN NRRNN NNRRN NNNRR -> 4 cases (=4!/3!)

2. 1 day in a row
a) 3 days of rain: RNRNR -> 1 case
b) 2 days of rain: RNRNN RNNRN RNNNR NRNRN NRNNR NNRNR -> 6 cases (=5!/3!2! - 4!/3!)
c) 1 days of rain: RNNNN NRNNN NNRNN NNNRN NNNNR -> 5 cases (=5!/4!)

3. 0 day in a row
a) 0 day of rain: NNNNN -> 1 case

The total outcome is 32. Therefore, the probability for the complimentary case is 24/32.

The probability that we want is 1 - 24/32 = 8/32, which also arrives at the correct answer!

My question is whether this approach is also valid? Or it is a mere coincidence?

Is LESS THAN 3 days in a row the complimentary case for AT LEAST 3 days in a row?
I notice that the permutation works for some cases (as written in the parentheses) and does not for many others. Could you please explain why?

Please help!

EMPOWERgmatRichC · Answer

Hi varotkorn,

When dealing with a Probability question, there are two outcomes that you can calculate: what you WANT to have happen or what you DON'T WANT to have happen. Those two fractions will sum to 1 (which is why you were able to calculate the correct answer by calculating the probability of NOT getting at least 3 days of rain in a row). Both methods are valid - and as a general rule, you should use whichever one is faster given the 'restrictions' in the prompt.

You made a reference to permutation calculations that did not "work" - can you give an example of a calculation that you tried that did not lead to the answer that you were expecting?

GMAT assassins aren't born, they're made, 
Rich

IanStewart · Answer

For most counting/probability questions, there's more than one way to solve. It's especially true in counting/probability that if you get the right answer, even using a method different from that in an 'official solution', your method is probably correct, because usually if you make a mistake, you either end up double-counting something or missing a case. I didn't check all of your work, but your method looks good to me.

In your post immediately above, where you're not getting the right answer, it looks like you're not excluding the cases where three R's occur in a row. As you're counting things, you're counting RR R N N as if it only has two Rs in a row, when it has three. You're also counting the words RR R N N and R RR N N as if they're different, but they're the same sequence. Instead you can count all the words you can make with R, R, R, N, N, which is 5! / (3!)(2!) = 10, then exclude the words with either no consecutive Rs (there's one of them, RNRNR) or with three consecutive Rs (there are three of those, RRRNN, NRRRN and NNRRR) to get 6.

KarishmaB · Answer

Nothing wrong with the complimentary case and hence you got the correct answer too. 
Though I wouldn't use this method because it is much easier to arrange fewer elements. When I glue together 3 Rs and must arrange them with 2 other elements, I just have to arrange 3 elements. 
But when I glue together 2 Rs and arrange the other 3 elements, I need to worry about 4 elements now. Similarly, 4Rs glued together means I just have to think about the 5th but with 1R in a row I have to ensure that there could be multiple Rs but none together - a far more complicated scenario.
Notice that in the straight forward case, you calculate 8 cases whereas in this complimentary case, you calculated 24 cases.

Kritisood · Answer

Hi rich, I tried your approach but aren't we getting 9 options in this? Am I missing something?

EMPOWERgmatRichC · Answer

Hi Kritisood,

Thanks for pointing this out. I went back and wrote in some additional explanation. In the original post, the sub-list in which 4 of the 5 days were rain-days included one option (RRNRR) that was NOT 3 days (or more) of rain - but it was not included in the total.
 
GMAT assassins aren't born, they're made,
Rich

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If the probability of rain on any given day is 50%, what is the probab

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